All the pairs (x, y) that satisfy the inequal | vedclass

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(B) Given inequality: $2^{\sqrt{\sin^2 x - 2\sin x + 5}} \cdot 4^{-\sin^2 y} \leq 1$Rewrite the inequality as: $2^{\sqrt{(\sin x - 1)^2 + 4}} \leq 4^{

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$\sin x = |\sin y|$

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(B) Given inequality: $2^{\sqrt{\sin^2 x - 2\sin x + 5}} \cdot 4^{-\sin^2 y} \leq 1$Rewrite the inequality as: $2^{\sqrt{(\sin x - 1)^2 + 4}} \leq 4^{\sin^2 y}$Since $4 = 2^2$,we have: $2^{\sqrt{(\sin x - 1)^2 + 4}} \leq 2^{2\sin^2 y}$Comparing the exponents: $\sqrt{(\sin x - 1)^2 + 4} \leq 2\sin^2 y$We know that $(\sin x - 1)^2 \geq 0$,so $\sqrt{(\sin x - 1)^2 + 4} \geq \sqrt{4} = 2$.Also,we know that $\sin^2 y \leq 1$,so $2\sin^2 y \leq 2$.For the inequality $2 \leq 	ext{LHS} \leq 	ext{RHS} \leq 2$ to hold,both sides must be equal to $2$.This implies $\sin x - 1 = 0 \Rightarrow \sin x = 1$ and $\sin^2 y = 1 \Rightarrow |\sin y| = 1$.Thus,$\sin x = 1$ and $|\sin y| = 1$,which means $\sin x = |\sin y|$.

All the pairs $(x, y)$ that satisfy the inequality $2^{\sqrt{\sin^2 x - 2\sin x + 5}} \cdot \frac{1}{4^{\sin^2 y}} \leq 1$ also satisfy the equation:

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